In the following problem, I want to find the angle marked as $x$. It seems so simple and yet I am out of ideas. It is very easy to get all angles except two of them: angle ADB and angle CBD.
Is there a calculation for the angle $x$ that only uses parallel lines and no circles?
Edit: A nice solution by using a circle is given below. In the book, this question was asked at the start of the chapter after introducing parallel lines and all the angles in that setup. Sum of angles of a triangle was proved and this question was asked in the exercise.
From geogebra, I know the answer is 60 degrees. But I do not know how to argue that the answer is 60 degrees.
Best Answer
Let $AD\cap BC=\{E\}$.
Thus, since $$\frac{AE}{CE}\cdot\frac{CE}{DE}\cdot\frac{DE}{BE}\cdot\frac{BE}{AE}=1$$ we obtain: $$\frac{\sin50^{\circ}}{\sin20^{\circ}}\cdot\frac{\sin60^{\circ}}{\sin50^{\circ}}\cdot\frac{\sin(70^{\circ}-x)}{\sin{x}}\cdot\frac{\sin80^{\circ}}{\sin30^{\circ}}=1$$ or $$\frac{\sin(70^{\circ}-x)}{\sin{x}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}.$$ But it's obvious that $x$ is an acute angle and $\frac{\sin(70^{\circ}-x)}{\sin{x}}$ decreases.
Id est, it's enough to prove that: $$\frac{\sin10^{\circ}}{\sin60^{\circ}}=\frac{\sin20^{\circ}\sin30^{\circ}}{\sin60^{\circ}\sin80^{\circ}}$$ or $$2\sin10^{\circ}\sin80^{\circ}=\sin20^{\circ},$$ which is obvious.